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               376 Examples, problems and exercises

                      (100 ‡ j0) ‡ j0:1   100e  j36:87    ˆ 106:3e j4:32    V. Note that the supply voltage E has
                      to be higher to achieve the same load voltage when the power-factor is lagging. The

                      load angle is d ˆ 4:32 and S ˆ VI ˆ 100   100e ‡j36:87    ˆ 8000 ‡ j6000 VA. Thus

                      S ˆ 10 kVA, P ˆ 8 kW and Q ˆ‡6 kVAr (absorbed).
                      Leading power-factor. The leading power-factor angle causes a reduction in the
                      value of E required to keep V constant: E ˆ 100 ‡ j0:1   100e ‡j36:87    ˆ 94:3e j4:86    V.
                      The load angle is d ˆ 4:86 , and S ˆ 10 000e  j36:87   ˆ 8000   j6000; i.e. P ˆ 8 kW and

                      Q ˆ 6 kVAr (generated).








                      Fig. 9.4 Leading PF.


                        We have seen that even though the power and the current are the same in all cases,
                      the inductive load with its lagging power factor requires a higher source voltage E.
                      The capacitive load with its leading power factor requires a lower source voltage.
                        If the source voltage E were kept constant, then the inductive load would have a
                      lower terminal voltage V and the capacitive load would have a higher terminal
                      voltage. As an exercise, repeat the calculations for E ˆ 100 V and determine V in
                      each case, assuming that Z ˆ 1 
 with each of the three different power-factors.
                        We can see from this that power-factor correction capacitors (connected in parallel
                      with an inductive load) will not only raise the power factor but will also increase the
                      voltage. On the other hand, if the voltage is too high, it could conceivably be reduced
                      by connecting inductors in parallel. In modern high-voltage power systems at loca-
                      tions far from the generating stations, it is possible to control the voltage by varying
                      the amount of inductive or capacitive current drawn from the system at the point
                      where the voltage needs to be adjusted. This is called reactive compensation or static
                      VAr control. In small, isolated power systems (such as an automotive or aircraft
                      power system supplied from one or two generators) this is not generally necessary
                      because the open-circuit voltage of the generator E can be varied by field control,
                      using a voltage regulator.



                        9.3   Simple basic problems

                      1. A single-phase power system has an open-circuit voltage E ˆ 6:35 kV and a fault
                        level of 16 MVA. Calculate the short-circuit current I sc in kA, and the The  venin
                        internal reactance X s in ohms. (See Figures 2.1 and 9.1).
                      2. What value of resistance R would draw 1.5 MW when connected to the power
                        system of Question 1? (See Figures 2.1 and 9.2). Calculate also the terminal
                        voltage V, the voltage drop X s I, and the load angle d.
                      3. What value of inductive reactance X L would be needed to reduce the voltage by
                        3% in the power system of Question 1 (assuming that no other loads are
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